Wrong. The body at 150 km altitude experiences a downward "force" due to
gravity ( personally prefer to think of it as curvature of space). Because
of its velocity, it is moving away from the earth even as the earth's
gravity moves it towards the earth. When these effects are in balance, we
say the object is "in orbit." The *net* force may be 0, but thet does not
mean that the individual forces are zero.

To solve the problem in old-fashioned Newtonian mechanics, the "force" of
gravity varies as 1/r^2 (from Newton's Law F = -G*M1*M2/r^2), so for two
given objects at two different distances (m1 and m2 are the same r changes)

F1 = -G*M1*M2/r1^2
F2 = -G*M1*M2/r2^2

so

F2/F1 = r1^2/r2^2

OTOH, the "force" with which the circular motion is throwing the smaller
object outwards is given by

Thanks for the clarification and I am sure what you have given me is very
useful but I am not sure what to do with it. (been about 9 years since I was
ina physics class)
I need a rearranged equation to give me the net effect of force in the y
plane as a projectile is travelling in a decaying orbit.
How do I calculate y force given speed and altitude ?

If the orbit is decaying, then there is something stealing angular momentum
and energy from the motion. This is called "friction" or "drag."

You need to know more, such as the drag coefficient. Given speed and
altitude you can calculate the energy and the angular momentum. Without
drag, the orbit will remain stable because angular momentum and energy are
conserved - i.e. they will not change. The motion and gravitation will stay
in balance.

To calculate the rate at which the object is "falling" - presumably the rate
at which the orbit is decaying, you need to know something else about what
is going on, such as what fraction of the energy or angular momentum is lost
in a given unit of time.